Two positively charged particles A and B of charges 'q' and ' 2 q ' respectively are released in a uniform electric field of intensity E. They are separated by some finite nonzero,

Question

Two positively charged particles A and B of charges 'q' and ' 2 q ' respectively are released in a uniform electric field of intensity E. They are separated by some finite nonzero, non negligible distance initially. At a later time $t = 10 s$ speed of 'B' is 'v' and that of 'A' is zero. Initially line joining A and B is parallel to the direction of electric field. What is the value of v in m/sec? (Neglect gravity)

TestHub · Accepted Answer

Impulse = Δ p ( q + 2 q ) E ⋅ t = m B ​ v ⇒ v = 1 3 × 1 0 − 6 × 1 0 5 ​ × 10 = 3 m / s

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